Question

1. Show that$\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$is a field.
2. Show that the field$\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$contains all three roots of${\mathbf{x}}^3+\mathbf{x}+1$

Short answer

1. 1. It is proved $\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$is a field.

Step-by-step solution

Statement of Theorem 5.10

Theorem 5.10states that, for the field and a nonconstant polynomial$p\left(x\right)$ in $F\left[x\right]$. The statements are equivalent as follows:

1. $p\left(x\right)$is irreducible in $F\left[x\right]$.
2. $F\left[x\right]/\left(p\left(x\right)\right)$willbe a field.
3. $F\left[x\right]/\left(p\left(x\right)\right)$will be an integral domain.

Show that ℤ2XX3+X+1 is a field

$\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$is a field because will be irreducible in ${\mathrm{\mathbb{Z}}}_2\left[X\right]$according to Theorem 5.10. It is observed that it will be a cubic polynomial with no root in ${\mathrm{\mathbb{Z}}}_2$.

It follows role="math" localid="1649249355584" $\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$is a field.

Hence, it is proved $\raisebox{1ex}{${\mathrm{\mathbb{Z}}}_2\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$\left(X^{3+}X+1\right)$}\right.$is a field.